$\begingroup$The launch rocket put gigajoules of energy into the satellite to accelerate it to orbital speed. One way or the other you need to dissipate that same energy to bring it back to 0 ground speed. The exact mechanism is a complex answer in the thermodynamics of gases, but the final answer is inevitable; one way or the other, a LOT of energy needs to be dissipated, typically as heat.$\endgroup$
– KengineerSep 14 '17 at 21:56

3 Answers
3

The term "friction" is a misnomer. The source of heat is adiabatic compression - gas on trajectory of the reentering object is compressed against its leading surface, and as result heats up.

On molecular level you can think of it as number of molecules rising in given volume (compressed) and additionally speeding up (by elastic collisions against the fast-moving surface). This all converts to massive temperature rise; also, these collisions constitute air drag, causing slowing down of the spacecraft.

Due to inertia of the gas, it takes some time before it moves sideways off the leading surface (giving away some of its heat to the object it touches at the time), and flies free off the edges, the following decompression (and resulting cooling) occurring far beyond the surface of the object, and so unable to cool it back down. There's also an adiabatic decompression on the trailing side, which would cool it down - except while the pressure there can drop only by 1 bar (from atmospheric to zero) the pressure can rise much more on the leading side, causing much more heating than cooling effect on the object.

At a certain point, the amount of heat is enough to turn any material of that temperature to plasma, thus the "flaming trail" as both the excited air and material of the body (be it rock of a meteorite, or ablator on reentering capsule) gets excited to the point of turning to plasma and blown away, leaving a blazing trail in the object's wake. Ablator - material that burns away carrying the heat with it, being a very poor heat conductor - is used on spacecraft to protect the inside of the craft from overheating through heat transfer from the superheated leading surface.

$\begingroup$Would a suction that pulls in the air from in from the leading surface and disperse it to the sides, at an equivalent speed as the decent so that compression never occurs allow an object to descend without incurring this phenomena?$\endgroup$
– PV22Aug 28 '18 at 18:51

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$\begingroup$@PV22 ...yes, except you won't get equivalent speed or anything remotely close. You can create 1 bar of suction (from ambient to 0) over an area that necessarily must be smaller than the suction device (so you won't cover the whole spacecraft) while you're getting possibly hundreds bars on the leading edge from the compression, all over the leading surface.$\endgroup$
– SF.Aug 28 '18 at 19:30

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$\begingroup$@PV22: The heating is in large part proportional to the relative speed of craft vs air (temperature is in essence speed of air particles; if particle hits a surface faster it passes more heat) so any extra motion would only worsen the situation (higher relative speed). Make the leading surface conical and indeed you'll distribute heat over larger surface and reduce the point heating. You'll also crash into the ground at Mach 15 or so. The reentry craft must be NON-aerodynamic!$\endgroup$
– SF.Aug 29 '18 at 12:51

$\begingroup$I had removed this because I didn’t want to other you, but... what about a wedge shaped nose or a conical coil in front to create turbulence in the air before the leading surface hits it?$\endgroup$
– PV22Aug 29 '18 at 14:24

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$\begingroup$@PV22: 1. what do you plan to make these things of, so that they don't burn up? 2. it's the same adiabatic compression that heats up the spacecraft and slows it down. Remove heating, you remove braking and crash at huge speed. You may at best move the heating somewhere else - e.g. from the ship's hull into the ablative coating of the heatshield. Remember each kilogram of the ship mass in orbit has 32 megajoules of kinetic energy while in orbit. You need to dissipate this whole energy before landing - as heat. That you heat ambient air outside (and not the crew cabin) is a good thing!$\endgroup$
– SF.Aug 29 '18 at 14:32

While @SF's answer is great in detailing how re-entry heating works, it doesn't directly address the question to the root causes, thus I have taken the time to answer the question you clarified in the comments:

Why do we need such a high re-entry velocity?

Orbital velocity in Low Earth Orbit (LEO) is ~7.8km/s. In order to re-enter, a short burn is performed at apogee to lower the perigee deep enough into the atmosphere, usually in the amount of a few dozen m/s up to a few hundred m/s. Still, the re-entry velocity will be higher than 7km/s. The Apollo capsules had more than 11km/s of velocity upon beginning of re-entry.

You can not slow down further, because of the so-called "Tyranny of the rocket equation". In order to slow down from 7.8 km/s to 0 m/s, you need a rocket that is roughly the same size as the rocket that launched you into orbit. And in order to launch that rocket into space you would need a rocket that is exponentially bigger.

Even if you could, having an object drop dead from 400km height isn't desirable. If you came to a full stop at ISS altitude somehow, you would be going ~2500m/s once passing the Karmann line, and ~3000m/s once at 50km height. but instead of having a rather long path through the atmosphere that slows the vessel down to terminal velocity, it would take the shortest possible path, and will likely not be able to deploy its chutes to slow down, resulting in lithobraking.

During SpaceX launches, the first stage reaches space and returns to earth without a heat shield. This is because the first stage does not achieve orbit, it stays on a suborbital trajectory and is much slower (from a quick web search, it seems to go about 2km/s at seperation), and thus re-entry heating is not a major concern. Add to that the fact that it is a powered landing, going down on quite a steep path is not that big of a problem - you don't need to be able to deploy chutes.

Finally re-entry heating is better than alternatives. When Apollo returned from the moon, they re-entered with more than 11km/s. This velocity could have been reduced by circularizing in LEO first, then re-entering. But the problem is again with orbital mechanics: You need the same amount of fuel for the trans-lunar-injection than you would need to circularize back around earth when coming back from the moon. This is quite a big amount of fuel, and not taking that fuel but instead a bigger heat shield is by far easier (again, tyranny of the rocket equation).

Returning interplanetary (e.g. from mars) will again have much higher re-entry speeds, and circularizing around earth won't be an option, either. You'd need as much delta-v to circularize around as you needed to leave earth to get to mars. It is simply not feasible to take such amounts of fuel on that journey.

So, you need to have high velocities during re-entry. And with those velocities, SF's answer sums up quite good what happens.

$\begingroup$just a note to any newbies, lithobraking is a humorous jargon of the space enthusiasts, derived from aerobraking where the atmosphere serves as the braking medium for the spacecraft. It's similar in case of lithobraking, except using the litosphere for delivering an extremely rapid loss of speed of the spacecraft (or, in other words, crashing.)$\endgroup$
– SF.Apr 27 '16 at 20:42

$\begingroup$Also, the trajectory of descent must be just right (luckily, in most cases that happens pretty much naturally). Too step, and the deceleration will kill/destroy everything or the craft will crash, not decelerating on time. Too shallow, and the deceleration will take too long, the heat will filter through the heatshield and the walls and fry everything and everyone inside. Very shallow (meaning not allowing overheating), and the descent would take weeks. At the right, moderately steep trajectory the craft slows to a safe speed before the heatshield leaks the heat inside.$\endgroup$
– SF.Apr 27 '16 at 20:47

$\begingroup$Lithobraking is not a humorous term - lithobraking is used in real spacecraft e.g. pathfinder and luna 9. Although when used for soft landings, it usually involves crumble zones and cushioning to prevent damage to the spacecraft. But it is good you mention the angle of descent.$\endgroup$
– PolygnomeApr 28 '16 at 11:40

Consider a typical craft that ascends into low Earth orbit then returns.

During ascent, the acceleration is provided by rocket thrust, spread over several minutes of flight, and the highest-speed portion of the flight is done above the dense part of Earth's atmosphere. Some heat is produced by compression (as SF's answer explains) and friction between the atmosphere and the spacecraft skin, but not a dramatic amount. At 45km altitude, the rocket may be going "only" mach 4 or 5, and air pressure is about 0.2% of sea level.

On reentry, no rocket thrust is slowing the craft -- all the deceleration will be done by atmospheric drag. When it first begins to enter the atmosphere, it will be going about mach 20; by the time it falls to 45km altitude, it will still be going around mach 10.

$\begingroup$why must there be a high speed of reentry?$\endgroup$
– PV22Apr 27 '16 at 1:43

$\begingroup$Because speed in low Earth orbit is over 7000 m/s, and decelerating with a rocket engine before entering the atmosphere would require a very large mass of propellant, which would necessitate a vastly larger launcher. The mass of a heat shield is much less.$\endgroup$
– Russell BorogoveApr 27 '16 at 1:54

$\begingroup$@alampert22: at lower speed the cushion of compressed air will be dissipating to the sides before achieving any significant pressure/temperature.$\endgroup$
– SF.Apr 27 '16 at 1:57

$\begingroup$Why do you need to be in a terminal orbit? Could you not approach from a direct trajectory?$\endgroup$
– PV22Apr 27 '16 at 1:59

$\begingroup$Direct trajectory from where? Any return from Moon or another planet would be even faster than LEO.$\endgroup$
– Russell BorogoveApr 27 '16 at 2:00